Question

What is the sum of the digits of the least number which, when divided by 52, leaves 33 as remainder; when divided by 78, leaves 59 and when divided by 117, leaves 98 as remainder?

Answer 1

 

My approach - Let number be X. 

So X= 52N+33=78M+59=117P+98, where M,N, P are divisors(?? I guess). 

We know that the smallest number which can be divided by 52, 78 and 117 is 13x2x2x3x3 = 52x9 = 468. 

Now, look at the remainders in each case, I can write 33 as (52-19), 59 as 78-19, and 98 as 117-19. Did you find a pattern. yes. 

Now rewrite the equation. 

X = 52(N+1) - 19 = 78(M+1) - 19 and 117(P+1) - 19. 

So now, the ques. boils down to find GCD of 52,78,and 117 ( which we know = 468 ) and deduct 19 from it = 449. 

Just to test the answer 449 / 52 = 9 + 33 remainder. 

449/78 = 5 + 59 remainder, and 449/117 = 3 + 98 remainder. 
So Answer is 449. 

2. A. It gains time. Instead coinciding every 65+5/11 mins. hands coincide every 65 mins. 

so gain of 5/11 every 65 mins.

Answer 2

The correct answer is 449.